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This question has probably been asked before on this website, but I could not find any solution and neither can I solve this question. So again I am asking the following question:

Let $\mathfrak{g}$ be a finite dimensional simple Lie algebra over $\mathbb{C}$ with Cartan subalgebra $\mathfrak{h}$ and $V$ be an irreducible weight module of $\mathfrak{g}$ with respect to $\mathfrak{h}$, i.e. $V= \oplus_{\lambda \in \mathfrak{h}^*} V_{\lambda}$ where $V_{\lambda} = \{v \in V : h.v = \lambda(h)v$ $\forall h \in \mathfrak{h}\}$. If $V_{\mu}$ is a non-zero finite dimensional weight space for some $\mu \in \mathfrak{h}^*$, then show that $V_{\lambda}$ is finite dimensional $\forall \lambda \in \mathfrak{h}^*$.

Any help or reference will be highly appreciated.

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    $\begingroup$ Yes, here is the last instance. The answer is affirmative and follows from the fact that $U_{\lambda}$ is a finitely generated left module over $U_0$, where the subscipts denote the grading of the universal enveloping algebra $U$ with respect to the adjoint action of the Cartan subalgebra. (There is too much notation for me to type a proper answer on a tablet.) $\endgroup$ – Victor Protsak Jul 19 at 23:42
  • $\begingroup$ I have been trying to prove that it is finitely generated for a number of days using the PBW theorem, but it has always evaded me. I know that proving your statement will give the desired result(though I needed the fact it is finitely generated as a right module). Can you please provide at least a sketch of the proof of your assertion here or in some other medium?Thank you. $\endgroup$ – Ester Jul 20 at 7:11
  • $\begingroup$ "Finite-dimensional simple Lie algebras" is not specific enough, I edited the title. $\endgroup$ – YCor Jul 20 at 15:52
  • $\begingroup$ No problem at all! $\endgroup$ – Ester Jul 20 at 15:58
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    $\begingroup$ Yes, we need finite generation of $U$ as a right $U_0$-module. The algebra $U$ is almost commutative (i.e. its associated graded algebra ${\rm gr\,}U$ is commutative and generated by degree 1 part) and the adjoint action of $\frak h$ is semisimple and preserves the filtration. So it is sufficient to prove the corresponding statement for the associated graded algebra: $S_{\lambda}$ is a finitely generated $S_0$-module. $\endgroup$ – Victor Protsak Jul 20 at 16:37
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The requisite property follows from the following key proposition:

$U_{\lambda}$ is a finitely generated right $U_0$-module.

Notation The subscripts denote the grading of the universal enveloping algebra $U=U({\frak g})$ with respect to the adjoint action of the Cartan subalgebra ${\frak h},\, \displaystyle U=\bigoplus_{\lambda\in P}U_{\lambda},$ with the grading abelian group the root lattice $P.$ The subspace $U_0$ is $U^{\frak h}$, the subalgebra of ${\frak h}$-invariants in $U.$ Similarly, $\displaystyle S=\bigoplus_{\lambda\in P}S_{\lambda}$ for the symmetric algebra $S=S({\frak g})$ and $S_0$ is the subalgebra of ${\frak h}$-invariants in $S.$

Proof of the property The action of $U_0$ stabilizes each weight subspace of $V$ and the action of $U_{\lambda}$ increases the weight by $\lambda$. Let $W=UV_{\mu}$, where a weight subspace $V_{\mu}$ is non-zero and finite-dimensional. Since $V$ is simple and $W$ is a non-zero submodule of $V$, $W=V$. Note that $U_{\lambda}V_{\mu}$ is $W_{\lambda+\mu}$, the weight subspace of $W$ of weight $\lambda+\mu$. Therefore $$\displaystyle V=\bigoplus_{\lambda\in P}U_{\lambda}V_{\mu}=\bigoplus_{\lambda\in P}W_{\lambda+\mu}$$ is the weight decomposition of $W=V$. For any $\lambda\in P$, $U_{\lambda}=X_{\lambda}U_0$ with a finite set $X_{\lambda}$, according to the proposition, and $U_{0}V_{\mu}=V_{\mu}$. It follows that each weight subspace $W_{\lambda+\mu}$ is finite-dimensional: $$W_{\lambda+\mu}=U_{\lambda}V_{\mu}=X_{\lambda}U_{0}V_{\mu}=X_{\lambda}V_{\mu}.$$

Proof of the proposition The algebra $U$ is almost commutative (i.e. its associated graded algebra ${\rm gr\,}U=S$ is commutative and generated by degree 1 part) and the adjoint action of $\frak h$ on $U$ is semisimple and preserves the filtration, so that ${\rm gr\,}U_{0}=S_{0}$ and ${\rm gr\,}U_{\lambda}=S_{\lambda}$. Hence it suffices to prove corresponding statement for the associated graded algebra:

$S_{\lambda}$ is a finitely generated $S_0$-module.

Recall that $S$ is a polynomial ring, the symmetric algebra of ${\frak g}$, and it is graded by $P$, i.e. it is a multigraded ring, and the last statement is a general property of multigraded rings. A good reference for these rings is the Miller-Sturmfels book "Combinatorial Commutative Algebra".

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  • $\begingroup$ I just have one simple question Sir. You said that S is graded by P, but as far as I know that S is graded by non-negative integers after taking the filtration of the universal enveloping algebra and then taking successive quotients. So can you please explain what do you mean by P grading here? I mean what is S_{\lambda} in your notation? Thank you. $\endgroup$ – Ester Jul 21 at 17:01
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    $\begingroup$ $S$ has two gradings, a positive grading by degree and a multigrading by $P$, as explained in Notation. The multigrading on $S$ comes from the weight decomposition of $\frak g$, i.e. the adjoint action of the Cartan subalgebra. $\endgroup$ – Victor Protsak Jul 21 at 17:05
  • $\begingroup$ Ok Sir. I shall certainly check. Thanks for your help and patience. $\endgroup$ – Ester Jul 21 at 17:06
  • $\begingroup$ Sir, I suppose you can put (gr(U)) within brackets in the last two instances where you have used them in your answer. I was terribly confused as I felt that you are referring to the grading on U rather than on gr(U). Thank you. $\endgroup$ – Ester Jul 22 at 9:13
  • $\begingroup$ Sir, the proof is completely clear to me except the last line where you say "Hence it suffices.....graded algebra."So can you please tell me how to conclude my assertion using $(grU)_0=S_0$ , $(grU)_{\lambda}=S_{\lambda}$ and the fact that $S_{\lambda}$ is finitely generated over $S_0$ ? Thank you. $\endgroup$ – Ester Jul 23 at 7:55
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EDIT: I misunderstood at first what your basic question is but now understand it better. One cautionary case comes from older work of Richard Block here, which includes the rank 1 simple Lie algebra and provides a classification of all irreducible representations of it (which had been regarded by Dixmier and others as an impossible task).

However, Block's classification and construction do not provide an immediate answer to your question. It is a natural case to begin with, but is already complicated to study.

In general there are plenty of irreducible $U(\mathbb{g})$-modules which are infinite dimensional but still in the BGG category $\mathcal{O}$. This category consists of modules (in characteristic 0) which satisfy several basic axioms including finite generation and being a direct sum of weight spaces. All modules in this category have finite dimensional weight spaces. Verma modules for example are generated by a one dimensional highest weight space.

The key property of the universal enveloping algebra of a semisimple Lie algebra is being noetherian and (via a PBW basis) having a nice "triangular" decomposition. Anyway, the concise 1976 paper (accessible online in the original Russian version) is one source, and my more leisurely but belated AMS textbook (and in later chapters survey) *Representations of Semisimple Lie Algebras in the BGG Category $\mathcal{O}$ (2008) gives a more comprehensive view of material including the Kazhdan-Lusztig conjecture of 1979 (soon a theorem). See especially Chapter 1.

P.S. Your auestion would be more clearly focused if you said "simple Lie algebra" in the header, or even "finite dimensional simple Lie algebra". This should be taken over any algebraically closed field such as $\mathbb{C}$ of characteristic 0.

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  • $\begingroup$ @JimHumphreys I don't think it's essential to have all the standing assumptions in the title (header), but that the title gives a quite good idea of the topic, and reasonably distinguishes the question within the topic (more specifically, say, than the tags). However such assumptions (finite-dimensional, over which field, which kind of representations, etc) certainly have to be specified in the question. $\endgroup$ – YCor Jul 20 at 16:11
  • $\begingroup$ Jim, she is not asserting that a simple weight module is necessarily finite-dimensional. The claim is that if a finite-dimensional simple module with a weight decomposition has one non-zero finite-dimensional weight space, then each weight space is finite-dimensional. There are plenty of simple weight modules beyond category $\mathcal O$ where this applies (classified by O.Mathieu and S.Fernando, see MR1775361). $\endgroup$ – Victor Protsak Jul 20 at 17:02
  • $\begingroup$ @Victor Just wanted to point out that it should be "he" rather than "she". $\endgroup$ – Ester Jul 20 at 17:05
  • $\begingroup$ @Ester: I thought at first you were working in the BGG category $\mathcal{O}$. where irreducible modules are to some extent known for a f.d. algebra. In general for a simple (or semisimple) Lie algebra these modules seem very difficult to study systematically beyond the examples considered by Matthieu and Fernando.. In category $\mathcal{O}$ BGG show your question has an affirmative answer for a semisimple finite dimensional Lie algebra. (Concerning the header, I'd still limit the setting to simple or semisimple Lie algebras, to avoid considering for example solvable ones.) $\endgroup$ – Jim Humphreys Jul 20 at 18:24
  • $\begingroup$ @Humphreys Thanks for the nice reference Sir. $\endgroup$ – Ester Jul 21 at 6:28

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